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1. Vai trò a, b, c như nhau. Không mất tính tổng quát. Giả sử \(a\ge b\ge0\)
Mà \(ab+bc+ca=3\). Do đó \(ab\ge1\)
Ta cần chứng minh rằng \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\left(1\right)\)
Và \(\frac{2}{1+ab}+\frac{1}{1+c^2}\ge\frac{3}{2}\left(2\right)\)
Thật vậy: \(\left(1\right)\Leftrightarrow\frac{1}{1+a^2}-\frac{1}{1+ab}+\frac{1}{1+b^2}-\frac{1}{1+ab}\ge0\\ \Leftrightarrow\left(ab-a^2\right)\left(1+b^2\right)+\left(ab-b^2\right)\left(1+a^2\right)\ge0\\ \Leftrightarrow\left(a-b\right)\left[-a\left(1+b^2\right)+b\left(1+a^2\right)\right]\ge0\\ \Leftrightarrow\left(a-b\right)^2\left(ab-1\right)\ge0\left(BĐT:đúng\right)\)
\(\left(2\right)\Leftrightarrow c^2+3-ab\ge3abc^2\\ \Leftrightarrow c^2+ca+bc\ge3abc^2\Leftrightarrow a+b+c\ge3abc\)
BĐT đúng, vì \(\left(a+b+c\right)^2>3\left(ab+bc+ca\right)=q\)
và \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\)
Nên \(a+b+c\ge3\ge3abc\)
Từ (1) và (2) ta có \(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\ge\frac{3}{2}\)
Dấu ''='' xảy ra \(\Leftrightarrow a=b=c=1\)
Áp dụng BĐT Cauchy dạng \(\frac{9}{x+y+z}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\), ta được
\(\frac{9}{a+3b+2c}=\frac{1}{a+c+b+c+2b}\le\frac{1}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)
Do đó ta được
\(\frac{ab}{a+3b+2c}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{a}{2}\right)\)
Hoàn toàn tương tự ta được
\(\frac{bc}{2a+b+3c}\le\frac{1}{9}\left(\frac{bc}{a+b}+\frac{bc}{b+c}+\frac{b}{2}\right);\frac{ac}{3a+2b+c}\le\frac{1}{9}\left(\frac{ac}{a+b}+\frac{ac}{b+c}+\frac{c}{2}\right)\)
Cộng theo vế các BĐT trên ta được
\(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\frac{1}{9}\left(\frac{ac+bc}{a+b}+\frac{ab+ac}{b+c}+\frac{bc+ab}{a+c}+\frac{a+b+c}{2}\right)=\frac{a+b+c}{6}\)Vậy BĐT đc CM
ĐẲng thức xảy ra khi và chỉ khi a = b = c >0
Với x, y > 0 ta chứng minh:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\\ \Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\\ \Leftrightarrow\left(x+y\right)^2\ge4xy\\ \Leftrightarrow\left(x-y\right)^2\ge0(luônđúng)\)
Dấu "=" xảy ra khi x = y
Áp dụng vào bài toán:
\(\frac{1}{a+b+2c}=\frac{1}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\ \Rightarrow\frac{4ab}{a+b+2c}\le\frac{ab}{a+c}+\frac{ab}{b+c}\)
Tương tự: \(\frac{4bc}{b+c+2a}\le\frac{bc}{b+a}+\frac{bc}{c+a}\\ \frac{4ca}{c+a+2b}\le\frac{ca}{c+b}+\frac{ca}{a+b}\\ 4\left(\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\right)\le\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ca}{a+b}=b+a+c\left(dpcm\right)\)
Dấu "=" xảy ra khi a = b = c
\(VT=\frac{a}{a+b+a+c}+\frac{b}{a+b+b+c}+\frac{c}{a+c+b+c}\)
\(VT\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{c}{b+c}\right)=\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c\)
Bài 2:
\(\frac{1}{\sqrt[3]{81}}\cdot P=\frac{1}{\sqrt[3]{9\cdot9\cdot\left(a+2b\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(b+2c\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(c+2a\right)}}\)
\(\ge\frac{3}{a+2b+9+9}+\frac{3}{b+2c+9+9}+\frac{3}{c+2a+9+9}\ge3\left(\frac{9}{3a+3b+3c+54}\right)=\frac{1}{3}\)
\(\Rightarrow P\ge\sqrt[3]{3}\)
Dấu bằng xẩy ra khi a=b=c=3
Bài 1:
\(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)
Theo bđt côsi-shaw ta luôn có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge\frac{25}{x+y+z+t+k}\)(x=y=z=t=k>0 ) (*)
\(\Leftrightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
Áp dụng bđt AM-GM ta có:
\(\hept{\begin{cases}x+y+z+t+k\ge5\sqrt[5]{xyztk}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge5\sqrt[5]{\frac{1}{xyztk}}\end{cases}}\)
\(\Rightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
\(\Rightarrow\)(*) luôn đúng
Từ (*) \(\Rightarrow\frac{1}{25}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\le\frac{1}{x+y+z+t+k}\)
Ta có: \(P=\frac{1}{2a+2b+c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\)
Mà \(\frac{1}{2a+2b+c}=\frac{1}{a+a+b+b+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{1}{a+2b+2c}=\frac{1}{a+b+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\frac{1}{2a+b+2c}=\frac{1}{a+a+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{25}\left[5.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=1\)
\(\Rightarrow P\le1\left(đpcm\right)\)Dấu"="xảy ra khi a=b=c\(=\frac{3}{5}\)
a) Dùng (a+b)2≥4ab
Chia hai vế cho a+b ( vì ab khác 0)
Ta có a+b≥\(\frac{4ab}{a+b}\) (Chuyển ab sang a+b) ta có
\(\frac{a+b}{ab}\)≥\(\frac{4}{a+b}\) <=> \(\frac{1}{a}\)+\(\frac{1}{b}\)≥\(\frac{4}{a+b}\)
Ta có:
\(\frac{a}{2a+b+c}=\frac{a}{\left(a+b\right)\left(a+c\right)}\le\frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
\(\frac{b}{a+2b+c}=\frac{b}{\left(a+b\right)\left(b+c\right)}\le\frac{b}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)
\(\frac{c}{a+b+2c}=\frac{c}{\left(a+c\right)\left(b+c\right)}\le\frac{c}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
Cộng vế theo vế:
=> \(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\le\frac{1}{4}\left(1+1+1\right)=\frac{3}{4}\)
Dấu "=" xảy ra <=> a = b = c
Cách 1:
Biến đổi tương đương bất đẳng thức cần chứng minh
\(1-\frac{a}{2b+b+c}+1-\frac{b}{a+2b+c}+1-\frac{c}{a+b+2c}\ge\frac{9}{4}\)
\(\Leftrightarrow\frac{a+b+c}{2a+b+c}+\frac{a+b+c}{a+2b+c}+\frac{a+b+c}{a+b+2c}\ge\frac{9}{4}\)
\(\Leftrightarrow4\left(a+b+c\right)\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\ge9\)
Đặt x=2a+b+c; y=a+2b+c; z=a+b+2c => x+y+z=4(a+b+c)
Khi đó đẳng thức trên trở thành
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-2\right)+\left(\frac{y}{z}+\frac{z}{y}-2\right)+\left(\frac{x}{z}+\frac{z}{x}-2\right)\ge0\)
\(\Leftrightarrow\frac{\left(x-y\right)^2}{2xy}+\frac{\left(y-z\right)^2}{2yz}+\frac{\left(z-x\right)^2}{2xz}\ge0\)
BĐT cuối luôn đúng
Vậy BĐT được chứng minh. Dấu "=" xảy ra <=> a=b=c
Cách 2:
Đặt x=2a+b+c; y=a+2b+c; z=a+b+2c
=> \(\hept{\begin{cases}a=\frac{2x-y-z}{4}\\b=\frac{3y-x-z}{4}\\c=\frac{3z-x-y}{4}\end{cases}}\)
BĐT cần chứng minh được viết lại thành
\(\frac{3x-y-z}{4x}+\frac{3y-x-z}{4y}+\frac{3z-x-z}{4z}\le\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{z}{x}\right)\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{z}{x}\ge6\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-2\right)+\left(\frac{y}{z}+\frac{z}{y}-2\right)+\left(\frac{x}{z}+\frac{z}{x}-2\right)\ge0\)
\(\Leftrightarrow\frac{\left(x-y\right)^2}{2xy}+\frac{\left(y-z\right)^2}{2yz}+\frac{\left(z-x\right)^2}{2zx}\ge0\)
BĐT cuối luôn đúng
Vậy BĐT được chứng minh. Dấu "=" <=> a=b=c
\(\frac{a}{b+c+2a}=\frac{a}{\left(a+b\right)+\left(a+c\right)}\le\frac{a}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}\)
Tương tự ta có:
\(\frac{b}{a+c+2b}\le\frac{b}{4\left(a+b\right)}+\frac{b}{4\left(b+c\right)}\)
\(\frac{c}{a+b+2c}\le\frac{c}{4\left(a+c\right)}+\frac{c}{4\left(b+c\right)}\)
Cộng vế theo vế ta có:
\(\frac{a}{b+c+2a}+\frac{b}{a+c+2b}+\frac{c}{a+b+2c}\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{c+b}+\frac{c}{a+c}+\frac{c}{b+c}\right)\)
\(=\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{a+c}{a+c}\right)=\frac{3}{4}\)
Dấu "=" xảy ra <=> a = b = c
Áp dụng bđt svacxo, ta có: \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\), ta có:
\(\frac{4}{2a+b+c}+\frac{4}{2b+c+a}+\frac{4}{2c+a+b}\le4\cdot\frac{\left(1+1+1\right)^2}{4\left(a+b+c\right)}=\frac{9}{a+b+c}\) (1)
Áp dụng bđt: \(\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\), ta có:
\(\frac{4}{2a+b+c}\le\frac{1}{a+b}+\frac{1}{a+c}\le\frac{1}{4}\cdot\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
CMTT:: \(\frac{4}{2b+c+a}\le\frac{1}{4}\cdot\left(\frac{2}{b}+\frac{1}{a}+\frac{1}{c}\right)\)
\(\frac{4}{2c+a+b}\le\frac{1}{4}\cdot\left(\frac{2}{c}+\frac{1}{a}+\frac{1}{b}\right)\)
=> \(\frac{4}{2a+b+c}+\frac{4}{2b+c+a}+\frac{4}{2c+a+b}\le\frac{1}{4}\cdot4.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) (2)
Từ (1) và (2)
=> \(\frac{9}{a+b+c}\le\frac{4}{2a+b+c}+\frac{4}{2b+c+a}+\frac{4}{2c+a+b}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Ta có: \(\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)
\(\ge\frac{\left(2+2+2\right)^2}{2a+b+c+a+2b+c+a+b+2c}\)
\(=\frac{6^2}{4\left(a+b+c\right)}=\frac{36}{4\left(a+b+c\right)}=\frac{9}{a+b+c}\)
Lại có áp dụng BĐT Cauchy - Schwarz ngược:
\(\frac{4}{2a+b+c}=\frac{1}{4}\cdot\frac{16}{a+a+b+c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{4}\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Tương tự:
\(\frac{4}{a+2b+c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{2}{b}+\frac{1}{c}\right)\) và \(\frac{4}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{2}{c}\right)\)
Cộng vế 3 BĐT trên lại ta được:
\(\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\le\frac{1}{4}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Dấu "=" xảy ra khi: a = b = c
Vậy \(\frac{9}{a+b+c}\le\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)